An infinite number of mathematicians walk into a bar. The first mathematician orders a beer. The second orders two beers. The next orders four. Followed by the next who orders eight. The bartender interrupts them: "ok guys, cut the crap: you owe me one beer".---------------------------------------------

A recently posted Numberphile video is heading towards 2 million hits on YouTube. That is an impressive score for a video focusing on a math subject. The two physicists in the video try to convince the viewer that all positive integers add up to -1/12. Yes, you read that correctly: these damn physicists dare to argue that 1 + 2 + 3 + 4 + ... = -1/12.

The video attracted some 5,000 reactions on YouTube including some dismissive reactions from several math purists. Phil Plait, the Bad Astronomer, who blogged enthusiastically about this Numberphile video attracted probably even more negative responses than the makers of the video themselves. Mark Chu-Carroll at Scientopia concluded"Phil Plait the Bad Astronomer, of all people, got taken in by a bit of mathematical stupidity, which he credulously swallowed and chose to stupidly expand on."

I do agree that the physicists in the Numberphile video oversimplify the math, yet the flurry of math-purist criticism dumped on the content of this video fails to attract my sympathy. Such "math rigor crusades" suggest the result 1 + 2 + 3 + 4 + ... = -1/12 to be erroneous, or at best to represent an utterly arbitrary and futile math exercise. This criticism diverts the attention away from some fascinating math that actually helps physicists taming divergences and carrying out meaningful calculations. Fact is that the infinite sum 1 + 2 + 3 + 4 + ... can be allocated only one value that has meaning, and that value is -1/12. Sure, you can maintain that this sum diverges and therefore can not be allocated a value, but as soon as you attempt allocating an actual value, there is no other meaningful choice than the choice -1/12.

Having said this, I do hold some reservations against the Numberphile video. Frankly, it discusses the wrong series. To convey intuitive arguments in favor of a unique negative value to be allocated to a never-ending sum of ever growing positive numbers, one has to select an example that allows the math to be kept simple and intuitive. The correct series to achieve this is not the sum of all positive integers, but rather the sum of all non-negative integer powers of two. The remarkable result to convey is that the divergent sum 1 + 2 + 4 + 8 + 16 + ... can be assigned the value minus unity.

This particular divergent sum has a long history, and the allocation of a finite value to this sum is covered by a dedicated Wikipedia page. Here I will zoom in on this sum with the objective to convey some simple arguments allowing you to build an intuition about regularizing divergent sums. Don't expect mathematical rigor from me: like the guys in the Numberphile video, I am a physicist raised in the time-honored tradition of 'catch-as-catch-can math' that laughs in the face of mathematical rigor.

Keeping this disclaimer in mind, we're ready to go.

Our first observation is that the value E to be assigned to the sum

E = 1 + 2 + 4 + 8 + 16 + ...

can not be chosen arbitrary. Some very basic requirements that most people would be willing to impose on recipes assigning values to sums (the requirements of linearity, stability and regularity to be discussed below), lead to E being uniquely defined. To see this, let's expand the sum such that it marches on in either direction:

S = ... + 1/16 + 1/8 + 1/4 + 1/2 + 1 + 2 + 4 + 8 + 16 + ...

What value S should we assign to this sum?

Whatever process we follow in assigning a value to a sum, the value assigned should be linear. For instance, if we double all terms in the sum, the sum itself should also double. Let's apply this to S:

So we have 2S = S. This equation has a single unique solution: S = 0. Of course you can maintain that S = infinite provides an alternative "solution". This alternative corresponds to the position that the series is divergent and can not be assigned a finite value. A perfectly valid position. However, the point here is that if one is prepared to assign a value to S, there is no choice: mathematical consistency is achieved only when S is set to zero.

This in itself represents a most remarkable observation. Any finite sum of positive numbers must be positive, yet we fail to observe the same for this infinite sum. Now let's focus on the difference between the sum

Demanding stability of values assigned to sums now leads to to the requirement

S = E + Z

We all know what value Z to assign to the Zeno sum. If you are presented a cake and you eat 1/2 of it, you are left with a remainder of 1/2 a cake. If you eat 1/2 of this remainder (1/4 cake), you are left with a 1/4 of a cake remainder. If again you eat 1/2 of this remainder (1/8 cake), you are left with 1/8 of the cake. Continuing this process, the remainder diminishes into ever tinier crumbs. When repeated ad infinitum, these crumbs vanish and you have eaten the full cake. So Z = 1, and regularity demands any credible summation method to reproduce this value.

We had already had established S = 0. The inevitable conclusion is that E = S - Z = -1:

1 + 2 + 4 + 8 + 16 + ... = -1

But how can this be? How can the addition of positive numbers roll over into negative?

At the end of the video, producer Brady Haran asks physicist Tony Padilla whether, if you kept adding integers forever on your calculator and hit the “equal” button at the end, you’d get -1/12. Padilla cheekily says, “You have to go to infinity, Brady!” But the answer should have been “No!”

It's a pity the video doesn't discuss the case of adding all powers of two, as in that case the answer would have been "Yes, a calculator would give -1". Obviously, you would need a powerful calculator that doesn't run into overflow issues. A 32-bit or 64-bit machine won't do the job. To handle a sum of powers of two that marches on forever, you need an "infinite-bit machine". Such a machine represents the sum of all non-negative powers of two in binary notation as an infinite string of 1's:

E = ...11111112

Let's assume we have stored this endless string of 1's in our infinite-bit machine. Next we make this infinite-bit machine add unity to this number. An infinite chain of carry-overs emerges that our infinitely powerful infinite-bit machine handles in a flash. The result is an infinite string of 0's:

Now suppose you are trying to predict the value of a certain physical parameter. You have available an approximation scheme that allows you to iterate towards ever more sophisticated approximations for this parameter. In zeroth order approximation you derive the value 1. The first order correction adds 2 to this value. The subsequent second order correction adds 4. Continuing this approximation scheme you derive a value 1 + 2 + 4 + 8 + 16 + ... for the parameter. This doesn't look too promising. You are losing hope you will ever be able to predict the value for the parameter. But you realize that the sum 1 + x + x2 + x3 + … converges to 1/(1−x) provided |x|<1. It seems your approximation method represents a clumsy attempt to calculate 1/(1-x) for x = 2. This means that the value you are seeking is 1/(1-2) = -1.

These are exactly the circumstances that make the regularisation of divergent sums such a useful branch of math. It allows physicists to tame infinity.

Comments

Sorry. I'm not convinced. If X = infinity and X + 1 = infinity then X = X + 1? But isn't there an infinite series of such combinations, each larger than its predecessor?The trick is implying that when you add 1 to your infinite series of bits that it will zero out -- but in fact it would never zero out (because it cannot reach a final bit), so it cannot be equal to zero.

Maybe generations of mathematicians know this stuff better than I can ever hope to, but it's still a mathematical parlor trick.

Zeno should have tried to explain what we should do with 1/2 of an infinite sum, for it surely must be less than infinite, right?

~The trick is implying that when you add 1 to your infinite series of bits that it will zero out -- but in fact it would never zero out (because it cannot reach a final bit), so it cannot be equal to zero.~ What particular bit(s) are you worried about? Which bit(s) would not flip to zero? Are you familiar with p-adics? Fascinating stuff.

So, what happens when we multiply the whole thing through by 2? Well, the series stays the same, so if (infinite series) = -1, then 2(infinite series) = 2(-1), but 2(infinite series) = (infinite series), so... the infinite series in this case = -2, and therefore -1 = -2. Which is why these tricks should *NOT* be considered valid math. It's like trying to get around the trap of 1/0 being indeterminant, by setting 1/0 = j (analogous to sqrt(-1) = i). Play around with it a bit, and you end up with -0 != +0, which sure seems problematic to me!

Why would I need to add 1? Ahhh, my bad. The series I was discussing was S, not E. I really should have phrased that reply better. Still, once an infinity is involved, the rest is just bullshit. And I'm not convinced Cantor was right, in the first place.

In which case E is NOT a progression, sum, series, or group or whatever magic thingamajig is supposed to make this felgercarb work. One of my old math professors said, "Infinity is not a real number so don't do real math with it." He left it at that. That always seemed fair to me.

I think the punch line is wrong - it should be "In exasperation and exhaustion the bartender says 'Hey guys, if you just give me back one beer, then you'll have enough for everyone and save me a lot of trouble!' ".